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2 years ago

Identify any incorrect formulas. Explain your answer.

1 answer:

5 0

A is incorrect because Mg has a 2+ charge, while SO4 has a 2- charge, if there is 2 Mg and 3 SO4 then the formula will not add to 0 meaning neutral

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Calculate the mass of Octane needed to release 6.20 mol Co2

n200080 [17]

The combustion reaction of octane is as follow,

C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O

According to balance equation,

8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted

So,

6.20 moles of CO₂ will release when = X g of Octane is reacted

Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol

X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.

8 0

2 years ago

Help please. Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1 g of potassium nitrate?

Masja [62]

Did u mean mass of potassium nitrate or the mass of nitrogen in potassium nitrate

6 0

3 years ago

A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, i

Alik [6]

Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

$X(s)+H_2O(l)\rightarrow X(aq)$

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

$m=1.00 g/mL\times 22.0 mL=22 g$

Mass of substance X = 2.50 g

Mass of the solution = M = 2.50 g + 22 g = 24.50 g

Heat released during the reaction be Q

Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C

Specific heat of the solution is equal to that of water :

c = 4.18J/(g°C)

$Q=Mc\times \Delta T$

$Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ$

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= $\frac{2.50 g}{82.0 g/mol}=0.03048 mol$

The enthalpy change, ΔH, for this reaction per mole of X:

$\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol$

5 0

2 years ago

PLEASE HELP ASAP !!

Ivenika [448]

Answer:

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Theoretical yield of H₂SO₄ = 213 g

percent yield of H₂SO₄ = 94 %

Explanation:

Data Given:

volume of SO₂ = 48.6 L

mass of H₂SO₄ = 200 g

balance equation = ?

theoretical yield = ?

percent yield = ?

Solution:

Part 1:

first we have to write a balance equation for the reaction

SO₂ gas react with water (H₂O) and excess oxygen

The balanced equation is as under

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Part 2:

Now we have to find theoretical yield

First look at the balance reaction

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

2 mol 2 mol

2 moles of SO₂ give gives 2 moles of H₂SO₄

Now calculate volume of 2 moles of SO₂ and mass of 2 moles of H₂SO₄

volume of 2 moles of SO₂

Formula used

volume of gas = no. of moles x molar volume . . . . . . (1)

molar volume of SO₂= 22.4 L/mol

Put values in above formula (1)

volume of gas = 2 mol x 22.4 L/mol

volume of gas = 44.8 L

volume of 2 mole of SO₂ = 44.8 L

Now,

Find mass of 2 mole H₂SO₄

Formula Used

mass in grams = no. of moles x molar mass . . . . . . . (2)

molar mass of H₂SO₄ = 2 (1) + 32 + 4(16)

molar mass of H₂SO₄ = 98 g/mol

put values in equation 2

mass in grams = 2 mol x 98 g/mol

mass in grams = 196 g

mass of 2 mole of H₂SO₄ = 196 g

** So,

Now we come to know that

44.8 L of SO₂ gives 196 g of H₂SO₄ then how many grams of the H₂SO₄ will be produced by 48.6 L of SO₂

Apply unity Formula

44.8 L of SO₂ ≅ 196 g of H₂SO₄

48.6 L of SO₂ ≅ X g of H₂SO₄

Do cross multiplication

g of H₂SO₄ = 196 g x 48.6 L / 44.8 L

g of H₂SO₄ = 213 g

So that is why the theoretical yield of H₂SO₄ is 213 g

Theoretical yield of H₂SO₄ = 213 g

Part 3

Calculate Percent Yield:

Formula used for this purpose:

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 200 g/ 213 g x 100 %

percent yield = 94 %

So percent yield of H₂SO₄ = 94 %

8 0

2 years ago

B. If the sand you ran across has a specific-heat capacity of 835 J/(kgºc),

Murrr4er [49]

Answer: 16700 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed = ?

m = mass of sand = 2 kg

c = heat capacity = $835J/kg^0C$

Initial temperature = $T_i$ = $40^0C$

Final temperature= $T_f$ = $50^0C$

Change in temperature , $\Delta T=T_f-T_i=(50-40)^0C=10^0C$

Putting in the values, we get:

$Q=2kg\times 835J/kg^0C\times 10^0C$

$Q=16700J$

16700 J of energy must be added to a 2-kilogram pile of it to increase its temperature from 40°C to 50°C

4 0

2 years ago